파이썬 | 백준 | 6593 | 상범 빌딩

solution

1. bfs이용 (6방향 → 3차원 배열)

# 6593, 상범빌딩
import sys
from collections import deque

dx = [-1, 1, 0, 0, 0, 0]
dy = [0, 0, -1, 1, 0, 0]
dz = [0, 0, 0, 0, -1, 1]


def bfs(z, x, y):
    visited[z][x][y] = 1
    q = deque()
    q.append((z, x, y))

    while q:
        c, a, b = q.popleft()
        for i in range(6):
            nz = c + dz[i]
            nx = a + dx[i]
            ny = b + dy[i]
            if 0 <= nz < L and 0 <= nx < R and 0 <= ny < C:
                if visited[nz][nx][ny] == 0 and sb[nz][nx][ny] != '#':
                    visited[nz][nx][ny] = visited[c][a][b] + 1
                    q.append((nz, nx, ny))
                    if sb[nz][nx][ny] == 'E':
                        return visited[c][a][b]
    return -1


while True:
    L, R, C = map(int, sys.stdin.readline().split())
    if L == 0 and R == 0 and C == 0:
        sys.exit()
    sb = []
    visited = [[[0] * C for _ in range(R)] for __ in range(L)]

    for z in range(L):
        b = []
        for x in range(R):
            b.append(list(sys.stdin.readline().rstrip()))
        sb.append(b)
        input()
    for z in range(L):
        for x in range(R):
            for y in range(C):
                if sb[z][x][y] == 'S':
                    res = bfs(z, x, y)
                    if res == -1:
                        print('Trapped!')
                    else:
                        print('Escaped in', res, 'minute(s).')

문제 출처 www.acmicpc.net/problem/6593